# Trigonometical identity proof

I was given a proving sum:

$\sec(x) + \tan (x) = p$, prove $\frac{p^2-1}{p^2+1} = \sin (x)$

I went head on and tried to directly do it by solving the LHS:

$\sec(x) + \tan(x)$ = $\frac{1}{\cos(x)} + \frac{\sin (x)}{\cos (x)}$ =$\frac{\sin(x)+1}{\cos(x)}$ Squaring both sides now,

= $(\sin(x)+1)^2 = p^2(\cos^2 (x))$

On further solving I am getting stuck. Can anyone point out how to proceed? I am curious to learn how to move beyond these steps. Also, easier answers (read: shorter steps) are always appreciated!

Just put the value of $p$ and simplify.

Use the facts : $\sec^2 x-1=\tan^2 x$ in numerator and $1+\tan^2 x=\sec^2 x$ in denominator.

$$\frac{p^2-1}{p^2+1}=\frac{2\tan x(\sec x+\tan x)}{2\sec x(\sec x+\tan x)}=\sin x$$

• Could you expand it please? I did not really get how you arrived at it directly? – Mission Coding Jul 27 '15 at 17:45
• Put $p$ and expand the square... – Empty Jul 27 '15 at 17:49
• Too much simple calculation...Do it yourself.. – Empty Jul 27 '15 at 17:52

$$\frac{p^2-1}{p^2+1} = 1 - \frac{2}{p^2+1}$$ You find that $$p=\frac{1+\sin x}{\cos x};$$ ok, then $$1 + p^2 = \frac{\cos^2x + (1+\sin x)^2}{\cos^2x} = \frac{\cos^2x + 1 + 2\sin x + \sin^2 x}{\cos^2x} = 2\frac{1 + \sin x}{\cos^2x}$$ and $$1 - \frac{2}{1+p^2} = 1 - \frac{\cos^2 x}{1 + \sin x} = \frac{1 - \cos^2x + \sin x}{1 + \sin x} = \frac{\sin2 x + \sin x}{1 + \sin x} = \sin x.$$

As $1=\sec^2x-\tan^2x=(\sec x-\tan x)(\sec x+\tan x),$

$\sec x+\tan x=p\iff\sec x-\tan x=\dfrac1p$

Now $\sin x=\dfrac{\tan x}{\sec x}$